Learning Objectives

Introduction

This lab was created by the developer of the R package prioritizr Jeffrey O. Hanson.

Overview

The aim of this workshop is to help you get started with using the prioritizr R package for systematic conservation planning. It is not designed to give you a comprehensive overview and you will not become an expert after completing this workshop. Instead, we want to help you understand the core principles of conservation planning and guide you through some of the common tasks involved with developing prioritizations. In other words, we want to give you the knowledge base and confidence needed to start applying systematic conservation planning to your own work.

Most importantly, the code needed to answer the questions in this workshop are almost always located in the same section as the question. So if you are stuck, try rereading the example code and see if you can modify it to answer the question.

Set Up

if (!require("librarian")){
  install.packages("librarian")
  library(librarian)
}
librarian::shelf(
  assertthat, 
  BiocManager, 
  dplyr, 
  gridExtra, 
  here, 
  mapview, 
  prioritizr, 
  prioritizrdata, 
  raster, 
  remotes, 
  rgeos, 
  rgdal, 
  scales, 
  sf, 
  sp, 
  stringr,
  units)

if (!require("lpsymphony")){
  BiocManager::install("lpsymphony")
  library(lpsymphony)
}

Download the data

The data for this workshop are available online. To download the data, please click on this link and save the data.zip file to your computer. Next, unzip the data.zip file. You should now have a new folder on your computer called "data" which contains the data files (e.g. pu.shp).

dir_data <- here("data/prioritizr")
pu_shp   <- file.path(dir_data, "pu.shp")
pu_url   <- "https://github.com/prioritizr/massey-workshop/raw/main/data.zip"
pu_zip   <- file.path(dir_data, basename(pu_url))
vegetation_tif <- file.path(dir_data, "vegetation.tif")

dir.create(dir_data, showWarnings = F, recursive = T)
if (!file.exists(pu_shp)){
  download.file(pu_url, pu_zip)
  unzip(pu_zip, exdir = dir_data)
  dir_unzip   <- file.path(dir_data, "data")
  files_unzip <- list.files(dir_unzip, full.names = T)
  file.rename(
    files_unzip, 
    files_unzip %>% str_replace("prioritizr/data", "prioritizr"))
  unlink(c(pu_zip, dir_unzip), recursive = T)
}

Data Import

# import planning unit data
pu_data <- as(read_sf(pu_shp), "Spatial")

# format columns in planning unit data
pu_data$locked_in <- as.logical(pu_data$locked_in)
pu_data$locked_out <- as.logical(pu_data$locked_out)

# import vegetation data
veg_data <- stack(vegetation_tif)

Planning unit data

The planning unit data contains spatial data describing the geometry for each planning unit and attribute data with information about each planning unit (e.g. cost values). Let’s investigate the pu_data object. The attribute data contains 5 columns with contain the following information:

id: unique identifiers for each planning unit
cost: acquisition cost values for each planning unit (millions of Australian dollars).
status: status information for each planning unit (only relevant with Marxan)
locked_in: logical values (i.e. TRUE/FALSE) indicating if planning units are covered by protected areas or not.
locked_out: logical values (i.e. TRUE/FALSE) indicating if planning units cannot be managed as a protected area because they contain are too degraded.

# print a short summary of the data
print(pu_data)

## class       : SpatialPolygonsDataFrame 
## features    : 516 
## extent      : 348703.2, 611932.4, 5167775, 5344516  (xmin, xmax, ymin, ymax)
## crs         : +proj=utm +zone=55 +south +datum=WGS84 +units=m +no_defs 
## variables   : 5
## names       :   id,             cost, status, locked_in, locked_out 
## min values  :  557, 3.59717531470679,      0,         0,          0 
## max values  : 1130,  47.238336402701,      2,         1,          1

# plot the planning unit data
plot(pu_data)

# plot an interactive map of the planning unit data
mapview(pu_data)

# print the structure of object
str(pu_data, max.level = 2)

## Formal class 'SpatialPolygonsDataFrame' [package "sp"] with 5 slots
##   ..@ data       :'data.frame':  516 obs. of  5 variables:
##   ..@ polygons   :List of 516
##   ..@ plotOrder  : int [1:516] 69 104 1 122 157 190 4 221 17 140 ...
##   ..@ bbox       : num [1:2, 1:2] 348703 5167775 611932 5344516
##   .. ..- attr(*, "dimnames")=List of 2
##   ..@ proj4string:Formal class 'CRS' [package "sp"] with 1 slot

# print the class of the object
class(pu_data)

## [1] "SpatialPolygonsDataFrame"
## attr(,"package")
## [1] "sp"

# print the slots of the object
slotNames(pu_data)

## [1] "data"        "polygons"    "plotOrder"   "bbox"        "proj4string"

# print the coordinate reference system
print(pu_data@proj4string)

## CRS arguments:
##  +proj=utm +zone=55 +south +datum=WGS84 +units=m +no_defs

# print number of planning units (geometries) in the data
row <- nrow(pu_data)

# print the first six rows in the data
head(pu_data@data)

##    id     cost status locked_in locked_out
## 1 557 29.74225      0     FALSE      FALSE
## 2 558 29.87703      0     FALSE      FALSE
## 3 574 28.60687      0     FALSE      FALSE
## 4 575 30.83416      0     FALSE      FALSE
## 5 576 38.75511      0     FALSE      FALSE
## 6 577 38.11618      2      TRUE      FALSE

# print the first six values in the cost column of the attribute data
head(pu_data$cost)

## [1] 29.74225 29.87703 28.60687 30.83416 38.75511 38.11618

# print the highest cost value
max_cost <- max(pu_data$cost)

# print the smallest cost value
min(pu_data$cost)

## [1] 3.597175

# print average cost value
mean(pu_data$cost)

## [1] 26.87393

# plot a map of the planning unit cost data
spplot(pu_data, "cost")

# plot an interactive map of the planning unit cost data
mapview(pu_data, zcol = "cost")

Questions:

How many planning units are in the planning unit data?

There are 516 planning units.

What is the highest cost value?

The highest cost value is 47.2383364.

Is there a spatial pattern in the planning unit cost values (hint: use plot to make a map)?

Based on the map, it seems like the eastern side of Tasmania has lower cost cells ranging from approximately 5 - 15. The middle and northern sections of the area of interest have the highest cost values hovering around 45. And the western side of Tasmania has middling cost values hoving around 30.

Vegetation data

The vegetation data describe the spatial distribution of 32 vegetation classes in the study area. This data is in a raster format and so the data are organized using a grid comprising square grid cells that are each the same size. In our case, the raster data contains multiple layers (also called “bands”) and each layer has corresponds to a spatial grid with exactly the same area and has exactly the same dimensionality (i.e. number of rows, columns, and cells). In this dataset, there are 32 different regular spatial grids layered on top of each other – with each layer corresponding to a different vegetation class – and each of these layers contains a grid with 164 rows, 326 columns, and 53464 cells. Within each layer, each cell corresponds to a 0.967 by 1.02 km square. The values associated with each grid cell indicate the (one) presence or (zero) absence of a given vegetation class in the cell.

Let’s explore the vegetation data.

# print a short summary of the data
print(veg_data)

## class      : RasterStack 
## dimensions : 164, 326, 53464, 32  (nrow, ncol, ncell, nlayers)
## resolution : 967, 1020  (x, y)
## extent     : 298636.7, 613878.7, 5167756, 5335036  (xmin, xmax, ymin, ymax)
## crs        : +proj=utm +zone=55 +south +datum=WGS84 +units=m +no_defs 
## names      : vegetation.1, vegetation.2, vegetation.3, vegetation.4, vegetation.5, vegetation.6, vegetation.7, vegetation.8, vegetation.9, vegetation.10, vegetation.11, vegetation.12, vegetation.13, vegetation.14, vegetation.15, ... 
## min values :            0,            0,            0,            0,            0,            0,            0,            0,            0,             0,             0,             0,             0,             0,             0, ... 
## max values :            1,            1,            1,            1,            1,            1,            1,            1,            1,             1,             1,             1,             1,             1,             1, ...

# plot a map of the 20th vegetation class
plot(veg_data[[20]])

# plot an interactive map of the 20th vegetation class
mapview(veg_data[[20]])

# print number of rows in the data
nrow(veg_data)

## [1] 164

# print number of columns  in the data
ncol(veg_data)

## [1] 326

# print number of cells in the data
ncell(veg_data)

## [1] 53464

# print number of layers in the data
nlayers(veg_data)

## [1] 32

# print  resolution on the x-axis
xres(veg_data)

## [1] 967

# print resolution on the y-axis
yres(veg_data)

## [1] 1020

# print spatial extent of the grid, i.e. coordinates for corners
extent(veg_data)

## class      : Extent 
## xmin       : 298636.7 
## xmax       : 613878.7 
## ymin       : 5167756 
## ymax       : 5335036

# print the coordinate reference system
print(veg_data@crs)

## CRS arguments:
##  +proj=utm +zone=55 +south +datum=WGS84 +units=m +no_defs

# print a summary of the first layer in the stack
print(veg_data[[1]])

## class      : RasterLayer 
## band       : 1  (of  32  bands)
## dimensions : 164, 326, 53464  (nrow, ncol, ncell)
## resolution : 967, 1020  (x, y)
## extent     : 298636.7, 613878.7, 5167756, 5335036  (xmin, xmax, ymin, ymax)
## crs        : +proj=utm +zone=55 +south +datum=WGS84 +units=m +no_defs 
## source     : vegetation.tif 
## names      : vegetation.1 
## values     : 0, 1  (min, max)

# print the value in the 800th cell in the first layer of the stack
print(veg_data[[1]][800])

##   
## 0

# print the value of the cell located in the 30th row and the 60th column of
# the first layer
print(veg_data[[1]][30, 60])

##   
## 0

# calculate the sum of all the cell values in the first layer
cellStats(veg_data[[1]], "sum")

## [1] 17

# calculate the maximum value of all the cell values in the first layer
cellStats(veg_data[[1]], "max")

## [1] 1

# calculate the minimum value of all the cell values in the first layer
cellStats(veg_data[[1]], "min")

## [1] 0

# calculate the mean value of all the cell values in the first layer
cellStats(veg_data[[1]], "mean")

## [1] 0.00035239

mapview(veg_data[[13]])

Questions:

What part of the study area is the 13th vegetation class found in (hint: make a map)? For instance, is it in the south-eastern part of the study area?

Most of the 13th vegetation class data are found in the eastern part of the study area.

What proportion of cells contain the 12th vegetation class?

total_cells <- ncell(veg_data)
twelfth_cells <- cellStats(veg_data[[12]], "sum")

twelfth_proportion <- twelfth_cells / total_cells
twelfth_percent <- round(twelfth_proportion * 100, digits = 2)

Approximately 1.53% of the cells contain the 12th vegetation class.

3. Which vegetation class is the most abundant (i.e. present in the greatest number of cells)?

Gap analysis

We need to understand how existing protected areas conserve native vegetation in Tasmania, Australia before we can prioritize areas for establishing protected areas
To do so, we will perform a “gap analysis” to assess how each biodiversity feature is represented by existing protected areas
Then we can compare the current representation to a target threshold of how much we want each feature represented
- target thresholds should be appropriate for how much area/habitat each species needs for ecosystem function and/or survival
- common values include 10% or 20%
- target thresholds can also be evaluated using policy or standard practices such as setting targets based on species geographic range size

Feature abundance

calculate how much of each vegetation feature occurs inside each planning unit
- this is the abundance of the features
use problem() to create an empty conservation planning problem p0 that only contains the planning unit and biodiversity data
- this assigns a cost value for each planning unit
use feature_abundances() to calculate the total amount of each feature in each planning unit
- this outputs a tibble, not a dataframe

# create prioritizr problem with only the PU data + vegetation data 
p0 <- problem(pu_data, veg_data, cost_column = "cost")

# print empty problem,
# we can see that only the cost and feature data are defined
print(p0)

# calculate amount of each feature in each planning unit
abundance_data <- feature_abundances(p0)

# print abundance data
print(abundance_data)

## # A tibble: 32 × 3
##    feature       absolute_abundance relative_abundance
##    <chr>                      <dbl>              <dbl>
##  1 vegetation.1                16.0                  1
##  2 vegetation.2                14.3                  1
##  3 vegetation.3                10.4                  1
##  4 vegetation.4                17.8                  1
##  5 vegetation.5                13.0                  1
##  6 vegetation.6                14.3                  1
##  7 vegetation.7                20.0                  1
##  8 vegetation.8                14.0                  1
##  9 vegetation.9                18.0                  1
## 10 vegetation.10               20.0                  1
## # … with 22 more rows

The abundance_data object contains three columns - The feature column contains the name of each feature (derived from names(veg_data)) - the absolute_abundance column contains the total amount of each feature in all the planning units - and the relative_abundance column contains the total amount of each feature in the planning units expressed as a proportion of the total amount in the underlying raster data. - since all the raster cells containing vegetation overlap with the planning units, all of the values in the relative_abundance column are equal to one (meaning 100%).

Add a new column with the feature abundances expressed in area units (i.e. km²).

# add new column with feature abundances in km^2
abundance_data$absolute_abundance_km2 <-
  (abundance_data$absolute_abundance * prod(res(veg_data))) %>%
  set_units(m^2) %>%
  set_units(km^2)

# print abundance data
print(abundance_data)

## # A tibble: 32 × 4
##    feature       absolute_abundance relative_abundance absolute_abundance_km2
##    <chr>                      <dbl>              <dbl>                 [km^2]
##  1 vegetation.1                16.0                  1                   15.8
##  2 vegetation.2                14.3                  1                   14.1
##  3 vegetation.3                10.4                  1                   10.2
##  4 vegetation.4                17.8                  1                   17.6
##  5 vegetation.5                13.0                  1                   12.8
##  6 vegetation.6                14.3                  1                   14.1
##  7 vegetation.7                20.0                  1                   19.7
##  8 vegetation.8                14.0                  1                   13.9
##  9 vegetation.9                18.0                  1                   17.8
## 10 vegetation.10               20.0                  1                   19.7
## # … with 22 more rows

Now let’s explore the abundance data.

# calculate the average abundance of the features
mean(abundance_data$absolute_abundance_km2)

## 86.82948 [km^2]

# plot histogram of the features' abundances
hist(abundance_data$absolute_abundance_km2, main = "Feature abundances")

# find the abundance of the feature with the largest abundance
max(abundance_data$absolute_abundance_km2)

## 737.982 [km^2]

# find the name of the feature with the largest abundance
abundance_data$feature[which.max(abundance_data$absolute_abundance_km2)]

## [1] "vegetation.12"

Now, try to answer the following questions.

What is the median abundance of the features (hint: median)?

median <- median(abundance_data$absolute_abundance_km2)

The median abundance of the features is 19.1165038783561 km².

What is the name of the feature with smallest abundance?

small <- abundance_data$feature[which.min(abundance_data$absolute_abundance_km2)]

The name of the feature with the smallest abundance is vegetation.3.

How many features have a total abundance greater than 100 km^2 (hint: use sum(abundance_data$absolute_abundance_km2 > set_units(threshold, km^2) with the correct threshold value)?

threshold = 100

total_abundance <- sum(abundance_data$absolute_abundance_km2 > set_units(threshold, km^2))

There are 6 features with a total abundance greater than 100 km².

Feature representation

calculate the amount of each feature in the planning units that are covered by protected areas (i.e. feature representation by protected areas)
use eval_feature_representation_summary(), which requires:
- 1. a conservation problem object with the planning unit and biodiversity data
- 1. an object representing a solution to the problem (i.e an object in the same format as the planning unit data with values indicating if the planning units are selected or not)
  - 1 = PU is covered by a protected area
  - 0 = PU is not covered by a protected area

# create column in planning unit data with binary values (zeros and ones)
# indicating if a planning unit is covered by protected areas or not
pu_data$pa_status <- as.numeric(pu_data$locked_in)

# calculate feature representation by protected areas
repr_data <- eval_feature_representation_summary(p0, pu_data[, "pa_status"])

# print feature representation data
print(repr_data)

## # A tibble: 32 × 5
##    summary feature       total_amount absolute_held relative_held
##    <chr>   <chr>                <dbl>         <dbl>         <dbl>
##  1 overall vegetation.1          16.0         0            0     
##  2 overall vegetation.2          14.3         0            0     
##  3 overall vegetation.3          10.4         0            0     
##  4 overall vegetation.4          17.8         0            0     
##  5 overall vegetation.5          13.0         0            0     
##  6 overall vegetation.6          14.3         0            0     
##  7 overall vegetation.7          20.0         0            0     
##  8 overall vegetation.8          14.0         0            0     
##  9 overall vegetation.9          18.0         0.846        0.0470
## 10 overall vegetation.10         20.0         0            0     
## # … with 22 more rows

The repr_data object contains three columns - The feature column contains the name of each feature - the absolute_held column shows the total amount of each feature held in the solution (i.e. the planning units covered by protected areas) - and the relative_held column shows the proportion of each feature held in the solution (i.e. the proportion of each feature’s spatial distribution held in protected areas).

the absolute_held values correspond to the number of grid cells in the veg_data object with overlap with protected areas
convert them to area units (i.e. km²) so we can report them.

# add new column with the areas represented in km^2
repr_data$absolute_held_km2 <-
  (repr_data$absolute_held * prod(res(veg_data))) %>%
  set_units(m^2) %>%
  set_units(km^2)

# print representation data
print(repr_data)

## # A tibble: 32 × 6
##    summary feature       total_amount absolute_held relative_held absolute_held_k…
##    <chr>   <chr>                <dbl>         <dbl>         <dbl>           [km^2]
##  1 overall vegetation.1          16.0         0            0                 0    
##  2 overall vegetation.2          14.3         0            0                 0    
##  3 overall vegetation.3          10.4         0            0                 0    
##  4 overall vegetation.4          17.8         0            0                 0    
##  5 overall vegetation.5          13.0         0            0                 0    
##  6 overall vegetation.6          14.3         0            0                 0    
##  7 overall vegetation.7          20.0         0            0                 0    
##  8 overall vegetation.8          14.0         0            0                 0    
##  9 overall vegetation.9          18.0         0.846        0.0470            0.834
## 10 overall vegetation.10         20.0         0            0                 0    
## # … with 22 more rows

Questions:

What is the average proportion of the features held in protected areas (hint: use mean(table$relative_held) with the correct table name)?

mean <- mean(repr_data$relative_held)
perc <- round(mean * 100, digits = 2)

On average, 24.15% of features are held in protected areas

If we set a target of 10% coverage by protected areas, how many features fail to meet this target (hint: use sum(table$relative_held >= target_value) with the correct table name)?

target_value = .10
sum <- sum(repr_data$relative_held >= target_value)

With a target of 10% coverage by protected areas, 15 features do not meet this target threshold.

If we set a target of 20% coverage by protected areas, how many features fail to meet this target?

target_value = .20
sum <- sum(repr_data$relative_held >= target_value)

With a target of 20% coverage by protected areas, 14 features do not meet this target threshold.

Is there a relationship between the total abundance of a feature and how well it is represented by protected areas (hint: plot(abundance_data$absolute_abundance ~ repr_data$relative_held))?

plot(abundance_data$absolute_abundance ~ repr_data$relative_held)

It seems like features that are more abundant typically are less represented by protected areas. This is what we might expect since protected areas usually prioritize rare or endangered species. Species with high total abundance values likely will not need to be protected.

Spatial prioritizations

develop prioritizations to identify priority areas for protected area establishment
prioritizr is a decision support tool (similar to Marxan and Zonation)
- it is designed to help you make decisions—it can’t make decisions for you.

Starting out simple

create a prioritization using the minimum set formulation of the reserve selection problem
we want a solution that will meet the targets for our biodiversity features for minimum cost
set 5% targets for each vegetation class
use the data in the cost column to specify acquisition costs

Although we strongly recommend using Gurobi to solve problems (with add_gurobi_solver), we will use the lpsymphony solver in this workshop since it is easier to install. - the Gurobi solver is much faster than the lpsymphony solver (see here for installation instructions).

# print planning unit data
print(pu_data)

## class       : SpatialPolygonsDataFrame 
## features    : 516 
## extent      : 348703.2, 611932.4, 5167775, 5344516  (xmin, xmax, ymin, ymax)
## crs         : +proj=utm +zone=55 +south +datum=WGS84 +units=m +no_defs 
## variables   : 6
## names       :   id,             cost, status, locked_in, locked_out, pa_status 
## min values  :  557, 3.59717531470679,      0,         0,          0,         0 
## max values  : 1130,  47.238336402701,      2,         1,          1,         1

# make prioritization problem
p1_rds <- file.path(dir_data, "p1.rds")
if (!file.exists(p1_rds)){
  p1 <- problem(pu_data, veg_data, cost_column = "cost") %>%
        add_min_set_objective() %>%
        add_relative_targets(0.05) %>% # 5% representation targets
        add_binary_decisions() %>%
        add_lpsymphony_solver()
  saveRDS(p1, p1_rds)
}
p1 <- readRDS(p1_rds)

# print problem
print(p1)

# solve problem
s1 <- solve(p1)

# print solution, the solution_1 column contains the solution values
# indicating if a planning unit is (1) selected or (0) not
print(s1)

## class       : SpatialPolygonsDataFrame 
## features    : 516 
## extent      : 348703.2, 611932.4, 5167775, 5344516  (xmin, xmax, ymin, ymax)
## crs         : +proj=utm +zone=55 +south +datum=WGS84 +units=m +no_defs 
## variables   : 7
## names       :   id,             cost, status, locked_in, locked_out, pa_status, solution_1 
## min values  :  557, 3.59717531470679,      0,         0,          0,         0,          0 
## max values  : 1130,  47.238336402701,      2,         1,          1,         1,          1

# calculate number of planning units selected in the prioritization
eval_n_summary(p1, s1[, "solution_1"])

## # A tibble: 1 × 2
##   summary  cost
##   <chr>   <dbl>
## 1 overall    15

# calculate total cost of the prioritization
eval_cost_summary(p1, s1[, "solution_1"])

## # A tibble: 1 × 2
##   summary  cost
##   <chr>   <dbl>
## 1 overall  385.

# plot solution
# selected = green, not selected = grey
spplot(s1, "solution_1", col.regions = c("grey80", "darkgreen"), main = "s1",
       colorkey = FALSE)

Questions:

How many planing units were selected in the prioritization? What proportion of planning units were selected in the prioritization?

n <- 15
N <- nrow(pu_data)

prop <- n / N
perc <- round(prop * 100, digits = 2)

The prioritization selected 15 planning units, which is 2.91 of the 516 total planning units.

Is there a pattern in the spatial distribution of the priority areas?

There is no clear pattern in the spatial distribution of the priority areas. They look scattered and random.

Can you verify that all of the targets were met in the prioritization (hint: eval_feature_representation_summary(p1, s1[, "solution_1"]))?

eval_feature_representation_summary(p1, s1[, "solution_1"])

## # A tibble: 32 × 5
##    summary feature       total_amount absolute_held relative_held
##    <chr>   <chr>                <dbl>         <dbl>         <dbl>
##  1 overall vegetation.1          16.0          2.90        0.181 
##  2 overall vegetation.2          14.3          1           0.0699
##  3 overall vegetation.3          10.4          1           0.0963
##  4 overall vegetation.4          17.8          3           0.168 
##  5 overall vegetation.5          13.0          1           0.0769
##  6 overall vegetation.6          14.3          1.94        0.136 
##  7 overall vegetation.7          20.0          1.75        0.0875
##  8 overall vegetation.8          14.0          2.80        0.200 
##  9 overall vegetation.9          18.0          2.16        0.120 
## 10 overall vegetation.10         20.0          2           0.0999
## # … with 22 more rows

Adding complexity

the first prioritization suffers many limitations
now, we should add additional constraints to the problem to make it more useful and realistic
lock in planning units that are already by covered protected areas
if some vegetation communities are already secured inside existing protected areas, then we might not need to add as many new protected areas to the existing protected area system to meet their targets
since our planning unit data (pu_da) already contains this information in the locked_in column, we can use this column name to specify which planning units should be locked in

# plot locked_in data
# TRUE = blue, FALSE = grey
spplot(pu_data, "locked_in", col.regions = c("grey80", "darkblue"),
       main = "locked_in", colorkey = FALSE)

# make prioritization problem
p2_rds <- file.path(dir_data, "p2.rds")

redo <- FALSE

if (!file.exists(p2_rds) | redo){
  p2 <- problem(pu_data, veg_data, cost_column = "cost") %>%
      add_min_set_objective() %>%
      add_relative_targets(0.05) %>%
      add_locked_in_constraints("locked_in") %>%
      add_binary_decisions() %>%
      add_lpsymphony_solver()
  saveRDS(p2, p2_rds)
}
p2 <- readRDS(p2_rds)

# print problem
print(p2)

# solve problem
s2 <- solve(p2)

# plot solution
# selected = green, not selected = grey
spplot(s2, "solution_1", col.regions = c("grey80", "darkgreen"), main = "s2",
       colorkey = FALSE)

Change target threshold to 10%

Let’s pretend that we talked to an expert on the vegetation communities in our study system and they recommended that a 10% target was needed for each vegetation class. So, equipped with this information, let’s set the targets to 10%.

# make prioritization problem
p3_rds <- file.path(dir_data, "p3.rds")
if (!file.exists(p3_rds)){
  p3 <- problem(pu_data, veg_data, cost_column = "cost") %>%
    add_min_set_objective() %>%
    add_relative_targets(0.1) %>%
    add_locked_in_constraints("locked_in") %>%
    add_binary_decisions() %>%
    add_lpsymphony_solver()
  saveRDS(p3, p3_rds)
}
p3 <- readRDS(p3_rds)

# print problem
print(p3)

# solve problem
s3 <- solve(p3)

# plot solution
# selected = green, not selected = grey
spplot(s3, "solution_1", col.regions = c("grey80", "darkgreen"), main = "s3",
       colorkey = FALSE)

Lock out highly degraded areas

Similar to before, this information is present in our planning unit data so we can use the locked_out column name to achieve this.

# plot locked_out data
# TRUE = red, FALSE = grey
spplot(pu_data, "locked_out", col.regions = c("grey80", "darkred"),
       main = "locked_out", colorkey = FALSE)

# make prioritization problem
p4_rds <- file.path(dir_data, "p4.rds")
if (!file.exists(p4_rds)){
  p4 <- problem(pu_data, veg_data, cost_column = "cost") %>%
    add_min_set_objective() %>%
    add_relative_targets(0.1) %>%
    add_locked_in_constraints("locked_in") %>%
    add_locked_out_constraints("locked_out") %>%
    add_binary_decisions() %>%
    add_lpsymphony_solver()
  saveRDS(p4, p4_rds)
}
p4 <- readRDS(p4_rds)

# print problem
print(p4)

# solve problem
s4 <- solve(p4)

# plot solution
# selected = green, not selected = grey
spplot(s4, "solution_1", col.regions = c("grey80", "darkgreen"), main = "s4",
       colorkey = FALSE)

Questions

What is the cost of the planning units selected in s2, s3, and s4?

# print solution, the solution_1 column contains the solution values - indicating if a planning unit is (1) selected or (0) not
print(s2)

## class       : SpatialPolygonsDataFrame 
## features    : 516 
## extent      : 348703.2, 611932.4, 5167775, 5344516  (xmin, xmax, ymin, ymax)
## crs         : +proj=utm +zone=55 +south +datum=WGS84 +units=m +no_defs 
## variables   : 7
## names       :   id,             cost, status, locked_in, locked_out, pa_status, solution_1 
## min values  :  557, 3.59717531470679,      0,         0,          0,         0,          0 
## max values  : 1130,  47.238336402701,      2,         1,          1,         1,          1

# calculate total cost of the prioritization
s2_cost <- eval_cost_summary(p2, s2[, "solution_1"])
s3_cost <- eval_cost_summary(p3, s3[, "solution_1"])
s4_cost <- eval_cost_summary(p4, s4[, "solution_1"])

The costs associated with s2, s3, and s4 are 6600.0886758, 6669.9087144, and 6711.581848 respectively.

How many planning units are in s2, s3, and s4?

# calculate number of planning units selected in the prioritization
s2_n <- eval_n_summary(p2, s2[, "solution_1"])
s3_n <- eval_n_summary(p3, s3[, "solution_1"])
s4_n <- eval_n_summary(p4, s4[, "solution_1"])

There are 205 planning units in s2, 211 planning units in s3, and 212 planning units in s4.

Do the solutions with more planning units have a greater cost? Why (or why not)?

Yes, s4 has the highest cost (6711.581848) with its overall, 212 planning units. Because solution 4 has the most constraints, it needs the most planning units to meet its 10% threshold. More planning units means there is a higher cost associated.

Why does the first solution (s1) cost less than the second solution with protected areas locked into the solution (s2)?

The first solution costs less than the second solution because in the second solution, we added the complexity of locking in existing protected areas. Thus, s2 has to select planning units from the remaining higher-cost area. The existing protected area likely covers low-cost planning units.

Why does the third solution (s3) cost less than the fourth solution solution with highly degraded areas locked out (s4)?

Penalizing fragmentation

plans for protected area systems should promote connectivity
the prioritizations we have made so far have been highly fragmented
we can add penalties to our conservation planning problem to penalize fragmentation
these penalties specify a trade-off between the primary objective (here, solution cost) and fragmentation (i.e. total exposed boundary length) using a penalty value
if we set the penalty value too low, then we will end up with a solution that is nearly identical to the previous solution
if we set the penalty value too high, then prioritizr will (1) take a long time to solve the problem and (2) we will end up with a solution that contains lots of extra planning units that are not needed.
the minimizing fragmentation is considered so much more important than solution cost that the optimal solution is simply to select as many planning units as possible
we generally want penalty values between 0.00001 and 0.01
ffinding a useful penalty value requires calibration
the “correct” penalty value depends on the size of the planning units, the main objective values (e.g. cost values), and the effect of fragmentation on biodiversity persistence.

Let’s create a new problem that is similar to our previous problem (p4)—except that it contains boundary length penalties—and solve it. Since our planning unit data is in a spatial format (i.e. vector or raster data), prioritizr can automatically calculate the boundary data for us.

# make prioritization problem
p5_rds <- file.path(dir_data, "p5.rds")
if (!file.exists(p5_rds)){
  p5 <- problem(pu_data, veg_data, cost_column = "cost") %>%
    add_min_set_objective() %>%
    add_boundary_penalties(penalty = 0.001) %>%
    add_relative_targets(0.1) %>%
    add_locked_in_constraints("locked_in") %>%
    add_locked_out_constraints("locked_out") %>%
    add_binary_decisions() %>%
    add_lpsymphony_solver()
  saveRDS(p5, p5_rds)
}
p5 <- readRDS(p5_rds)

# print problem
print(p5)

# solve problem,
# note this will take a bit longer than the previous runs
s5 <- solve(p5)

# print solution
print(s5)

## class       : SpatialPolygonsDataFrame 
## features    : 516 
## extent      : 348703.2, 611932.4, 5167775, 5344516  (xmin, xmax, ymin, ymax)
## crs         : +proj=utm +zone=55 +south +datum=WGS84 +units=m +no_defs 
## variables   : 7
## names       :   id,             cost, status, locked_in, locked_out, pa_status, solution_1 
## min values  :  557, 3.59717531470679,      0,         0,          0,         0,          0 
## max values  : 1130,  47.238336402701,      2,         1,          1,         1,          1

# plot solution
# selected = green, not selected = grey
spplot(s5, "solution_1", col.regions = c("grey80", "darkgreen"), main = "s5",
       colorkey = FALSE)

Questions:

Now let’s compare the solutions to the problems with (s5) and without (s4) the boundary length penalties.

What is the cost the fourth (s4) and fifth (s5) solutions? Why does the fifth solution (s5) cost more than the fourth (s4) solution?

# calculate total cost of the prioritization
s4_cost <- eval_cost_summary(p4, s4[, "solution_1"])
s5_cost <- eval_cost_summary(p5, s5[, "solution_1"])

The fourth solution costs 6711.581848, and the fifth solution costs 6747.585313. The fifth solution costs more because of the boundary penalty we placed on it. The boundary penalty forced the solution to clump planning units together even if those planning units were relatively high cost.

Try setting the penalty value to 0.000000001 (i.e. 1e-9) instead of 0.001. What is the cost of the solution now? Is it different from the fourth solution (s4) (hint: try plotting the solutions to visualize them)? Is this is a useful penalty value? Why (or why not)?

# make prioritization problem
p6_rds <- file.path(dir_data, "p6.rds")
if (!file.exists(p6_rds)){
  p6 <- problem(pu_data, veg_data, cost_column = "cost") %>%
    add_min_set_objective() %>%
    add_boundary_penalties(penalty = 0.000000001) %>%
    add_relative_targets(0.1) %>%
    add_locked_in_constraints("locked_in") %>%
    add_locked_out_constraints("locked_out") %>%
    add_binary_decisions() %>%
    add_lpsymphony_solver()
  saveRDS(p6, p6_rds)
}
p6 <- readRDS(p6_rds)

# print problem
print(p6)

# solve problem,
# note this will take a bit longer than the previous runs
s6 <- solve(p6)

# print solution
print(s6)

## class       : SpatialPolygonsDataFrame 
## features    : 516 
## extent      : 348703.2, 611932.4, 5167775, 5344516  (xmin, xmax, ymin, ymax)
## crs         : +proj=utm +zone=55 +south +datum=WGS84 +units=m +no_defs 
## variables   : 7
## names       :   id,             cost, status, locked_in, locked_out, pa_status, solution_1 
## min values  :  557, 3.59717531470679,      0,         0,          0,         0,          0 
## max values  : 1130,  47.238336402701,      2,         1,          1,         1,          1

# plot solution
# selected = green, not selected = grey
spplot(s6, "solution_1", col.regions = c("grey80", "darkgreen"), main = "s6",
       colorkey = FALSE)

s6_cost <- eval_cost_summary(p6, s6[, "solution_1"])

The new solution with a very low penalty cost is 6711.7559287, which is very similar to the solution 4 cost of 6711.581848. The two maps also look identical, with lots of fragments in the eastern side of Tasmania. Since the two solutions look very similar, this was not a helpful penalty value.

Try setting the penalty value to 0.5. What is the cost of the solution now? Is it different from the fourth solution (s4) (hint: try plotting the solutions to visualize them)? Is this a useful penalty value? Why (or why not)?

p7_rds <- file.path(dir_data, "p7.rds")
if (!file.exists(p7_rds)){
  p7 <- problem(pu_data, veg_data, cost_column = "cost") %>%
    add_min_set_objective() %>%
    add_boundary_penalties(penalty = 0.5) %>%
    add_relative_targets(0.1) %>%
    add_locked_in_constraints("locked_in") %>%
    add_locked_out_constraints("locked_out") %>%
    add_binary_decisions() %>%
    add_lpsymphony_solver()
  saveRDS(p7, p7_rds)
}
p7 <- readRDS(p7_rds)

# print problem
print(p7)

# solve problem,
# note this will take a bit longer than the previous runs
s7 <- solve(p7)

# print solution
print(s7)

## class       : SpatialPolygonsDataFrame 
## features    : 516 
## extent      : 348703.2, 611932.4, 5167775, 5344516  (xmin, xmax, ymin, ymax)
## crs         : +proj=utm +zone=55 +south +datum=WGS84 +units=m +no_defs 
## variables   : 7
## names       :   id,             cost, status, locked_in, locked_out, pa_status, solution_1 
## min values  :  557, 3.59717531470679,      0,         0,          0,         0,          0 
## max values  : 1130,  47.238336402701,      2,         1,          1,         1,          1

# plot solution
# selected = green, not selected = grey
spplot(s7, "solution_1", col.regions = c("grey80", "darkgreen"), main = "s7",
       colorkey = FALSE)

s7_cost <- eval_cost_summary(p7, s7[, "solution_1"])

The final solution has a cost of 9816.639992, which is much higher than the s4 cost of 6711.581848. This is because the extremely high penalty cost forced the solution to lump every single planning cell together, even extremely high cost cells. This is the opposite end of the spectrum of s6 and still too extreme to be useful for conservation planning.

Lab 3: Reserve Planning

Mia Forsline

2022-02-02

Learning Objectives

Introduction

Overview

Set Up

Download the data

Data Import

Planning unit data

Questions:

Vegetation data

Questions:

3. Which vegetation class is the most abundant (i.e. present in the greatest number of cells)?

Gap analysis

Feature abundance

Feature representation

Questions:

Spatial prioritizations

Starting out simple

Questions:

Adding complexity

Change target threshold to 10%

Lock out highly degraded areas

Questions

Penalizing fragmentation

Questions: